Pre Calculus Polar Equations & Functions

In this lesson, we are going to talk about polar equations and functions. We can set up equations and functions using r and θ, exactly the same way as we did for x and y. Generally, θ is the independent variable (like x was), while r is the dependent variable (like y was). We graph polar equations/functions in the same way we graph normal stuff: plug in values, plot points, and connect with curves to make a graph. Since θ is the independent variable, we plug in some value for it, then see what distance r we get out at that angle. You'll learn some tips for graphing and converting coordinate types.

Polar Equations & Functions

We can set up equations and functions using r and θ exactly the same way as we did for x and y. Generally, θ is the independent variable (like x was), while r is the dependent variable (like y was).

We always assume that θ is in radians ([(π)/2], [(7π)/4], etc.).

We graph polar equations/functions in the same way we graph "normal" stuff: plug in values, plot points, and connect with curves to make a graph. Since θ is the independent variable, we plug in some value for it, then see what distance r we get out at that angle.

Just like graphing with rectangular equations, you don't need to plot a huge number of points-merely enough to sketch the graph. Many polar equations involve trigonometric functions. The "interesting" points are when the trig function produces a zero or an extreme value (sin, cos ⇒ ±1). Figure out where these interesting values will occur and use them to help plot the graph.

If at anytime you're unsure how the graph will behave, just plot more points. The easiest way to work out your uncertainty is by calculating more points.

Occasionally you will see an equation that only uses one variable: that's fine! It means that variable is fixed, while the other can change freely.

Sometimes we'll want to convert an entire equation or function from polar to rectangular, or vice-versa. We can do so with the same conversion formulas we figured out and used in the previous lesson. Since these formulas were based off any x, y, r, θ, they work fine for substituting in equations.

Polar ⇒ Rectangular: x = r cosθ y = rsinθ

Rectangular ⇒ Polar: r2 = x2 + y2 tanθ = y/x

If you have access to a graphing calculator, it's great to try graphing some polar equations with it. It's a new way of looking at graphing, so it helps just to play around. For more information, check out the appendix on graphing calculators.

Polar Equations & Functions

Fill in the table of values for the polar function below.

r(θ) = 3+cos(2θ)

θ

r

0

π

4

π

2

3π

4

π

5π

4

3π

2

7π

4

2π

Filling in a table of values for a polar function is the same as doing it for a normal function. The input value is θ for the function, so we plug in some value for θ, then we see what output value we get for r.

For example, to find r(0), we plug in like normal:

r(0) = 3 + cos(2 ·0) = 3 + cos(0) = 3+1 = 4

Repeat the above process for each value of θ in the table:

θ

r

0

4

π

4

3

π

2

2

3π

4

3

π

4

5π

4

3

3π

2

2

7π

4

3

2π

4

[Notice that the values of r repeat after θ = π. This is because cos(2θ) has a period of π. Still, the values in the table after that are important because the θ values are distinctly different. When it comes time to draw polar graphs, remember that the θ value is just as key to plotting a point as the r value.]

θ

r

0

4

[(π)/4]

3

[(π)/2]

2

[(3π)/4]

3

π

4

[(5π)/4]

3

[(3π)/2]

2

[(7π)/4]

3

2π

4

Below is the graph of r(θ) = 3cos(5 θ). Find an interval of θ where the graph is only drawn out once. (That is, the graph does not get re-traced as later values for θ wind up creating matching locations to those already drawn in by earlier values for θ.)

In polar coordinates, it's possible (and common) for numerically different points to give the same location. For example, all of the below points are equivalent:

(7, π) ≡ (7, 3π) ≡ (−7, 0)

[If this idea is unfamiliar to you, make sure to check out the previous lesson on the idea of polar coordinates. Being comfortable with this idea and how polar coordinates are plotted is critical to understanding polar equations/functions.]

We are looking for when the equation r=3 cos(5θ) begins to repeat locations. While it will never repeat the same coordinate numerically, it will eventually create equivalent coordinates, and the equation will re-trace the locations the graph has covered in previous points. Let us arbitrarily choose a starting angle of θ = 0. We could start before that, but 0 is a nice, easy place to start. With that in mind, notice that we will be pointing back in the same direction at θ = 2π. Thus, if we get the same r value for θ = 0 and θ = 2π, we see that the graph will be repeating then. Plug in to check:

r(0) = 3cos(5 ·0) = 3

⎢⎢

r(2π) = 3cos(5 ·2π) = 3

Furthermore, if we continue onward from θ = 2π, all the values will be the exact same as if we had started at θ = 0 because of the periodic nature of the trig function. Thus, the graph is repeating after 2π. However, there is one more thing to keep in mind...

It is also possible to have duplicate locations from different coordinates if the r is negative in one of them while the θ points in the opposite direction. For example, (7, π) ≡ (−7, 2π). This means that, assuming we start at θ = 0, there is another, earlier opportunity for repetition: θ = π. If that causes the function to spit out a negative version of the r, they will be giving equivalent locations:

r(0) = 3cos(5 ·0) = 3

⎢⎢

r(π) = 3cos(5 ·π) = −3

Thus we see that, yes, r(0) and r(π) graph to the exact same location. Furthermore, if we look at how r(θ) will behave for angles starting at θ = π, we see that it will produce the exact opposite (negative) to what it produced when starting at θ = 0. [This happens because θ = π catches cos(5θ) in the exact middle of a period interval, so everything is flipped to the opposite from there on out.] Thus, we see that when we start at θ = 0, the first repetition of the function begins at θ = π. This means we can create the entire graph of the function by just using the interval θ: [0, π).

Finally, if you found all the above a little difficult to follow, don't worry! Parametric coordinates are confusing when they're new to you, and parametric equations/functions are even tougher the first time around. A great way to help yourself understand what's happening is to use a graphing calculator. Enter in the function to be graphed (don't forget to give it an interval of θ to graph, most graphing calculators will require you to enter one), then try out the trace function. Trace the graph using the calculator and get a better understanding of how it's working. Feel free to play around with the functions you're looking at too! Adjust some numbers, try other trig functions, and just generally explore. The easiest way to get comfortable with polar equations is by exploring them, and a graphing calculator makes it possible to see a lot of different things very quickly.

θ: [0, π)

Graph the polar equation: r = sin(θ).

Unless you're already extremely familiar with a certain type of equation, it's almost always necessary to create a table of values to help see how the function graphs. Make sure to get enough values so that you understand what's going on in the equation and can comfortably graph it. If you're ever unsure about what it will look like, just plot more points until it becomes clear to you.

We might create a table of values like the one below:

θ

r

0

0

π

4

0.71

π

2

1

3π

4

0.71

π

0

5π

4

−0.71

3π

2

−1

7π

4

−0.71

2π

0

Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice how sin(θ) hits its maximum at θ = [(π)/2]. It starts at 0, then increases to 1 as it rotates, then comes back down to 0. The graph is then re-traced by r becoming negative, telling us to go in the opposite direction to the θ angle, causing us to cover the same graph again.

Graph the polar equation θ = − [(π)/6].

Notice that this polar equation has no r term in it whatsoever. The only thing that has a restriction on it is the θ term.

This means that θ is restricted to only being −[(π)/6], but that r can be anything at all (since r does not appear anywhere, the equation puts no restrictions on it). Thus, we set the angle of θ = −[(π)/6] while r is allowed to run the interval of (−∞, ∞). This is similar to a rectangular equation like x=5. We set x as 5, then y is allowed to run anywhere, so we get a vertical line out of that equation.

When drawing in the graph, don't forget that negative values for r cause them to go "backwards". This means θ sets an angle to point in, and the various possible r values fill out the forward and backward directions for that angle, creating a line.

Graph the polar equation r=√{θ} with the restriction that θ ≤ 4 π.

Begin by noticing that while there is an explicit upper restriction on what θ can be (the problem says θ ≤ 4 π), there is also an implied lower restriction. The equation we're working with is r=√{θ}: if θ < 0, then the equation won't make any sense, because we can't take the square root of negative. Thus, the only domain we have to consider for this problem is θ: [0, 4π].

Once we know where we're looking in terms of θ, we approach the problem like other graphing problems. Create a table of values so you have a sense of how r grows and changes and so you can later plot points. Using a calculator to find approximate values for r, we have:

θ

r

0

0

π

2

1.25

π

1.77

3π

2

2.17

2π

2.51

5π

2

2.80

3π

3.07

7π

2

3.32

4π

3.54

As always, if you're unsure about how the graph works or would like to have more plots to point, just calculate some extra points until you feel comfortable with what you have.

Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice that r=√{θ} grows very quickly at first, but slows down the larger it gets. This means that the changes when it first starts out at θ = 0 will be especially quick, so it would probably be useful to get a few extra points for those early θ-values:

θ

r

π

6

0.72

π

4

0.88

π

3

1.02

This goes to show just how quickly r grows for early values of θ. Although r=0 when θ = 0, it shoots out very quickly for the first tiny bit of turning. As θ turns more, the growth rate of r slows down.

Graph the polar function r(θ) = 1+4sin(2θ).

Unless you're already extremely familiar with a certain type of equation, it's almost always necessary to create a table of values to help see how the function graphs. Make sure to get enough values so that you understand what's going on in the equation and can comfortably graph it. If you're ever unsure about what it will look like, just plot more points until it becomes clear to you. When figuring out what values of θ to use in your table, think in terms of which values will be "interesting". For example, the most "interesting" values for cos(x) are x=0, [(π)/2], π, [(3π)/2]. These values are extremely important because they make up the zeros, maximum, and minimum for cos(x). Along these lines, think of what values for θ will be "interesting" for sin(2θ).

Because sin(2θ) effectively goes twice as "fast" as sin(θ), this causes all the interesting values to occur on intervals of [(π)/4]. We will see our zeros, maximums, and minimums all fall on the below values:

θ = 0,

π

4

,

π

2

,

3π

4

, …

Furthermore, notice that sin(2θ) will start repeating outputs after π since π is its period. We still have to care about the θ-values after π because, even though the r-values will repeat, the θ-values will indicate new angles and thus create unique locations.

θ

r

0

1

π

4

5

π

2

1

3π

4

−3

π

1

5π

4

5

3π

2

1

7π

4

−3

2π

1

As always, if you're unsure about how the graph works or would like to have more points to plot, just calculate some extra points until you feel comfortable with what you have.

Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice that we get the exact same r values, in the same order, for θ from [0, π] as we do for [π, 2 π]. This, combined with how the angle is spinning, causes the graph to mirror around the origin. Noticing this mirroring effect can make it easier to draw in the graph. As you get really skilled, you can even notice this sort of mirroring before ever making the table, allowing you to create graphs much more quickly but without losing any quality.

Convert the equation from polar to rectangular.

r = cos(θ)

We can convert from polar coordinates to rectangular coordinates through the use of the following identities:

x = rcosθ, y = rsinθ, r2 = x2 + y2, tanθ =

y

x

When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.

With this idea in mind, we see that it's difficult to directly swap in an identity for the equation r = cos(θ): neither side directly fits one of the identities. However, we see that we're close to an identity. If we had r2 or r cosθ, we could match up to an identity. We can achieve both of these by multiplying each side of the equation by r:

r = cos(θ) ⇒ r ·r = r·cos(θ) ⇒ r2 = r cos(θ)

With this new formation of the original equation, it's now quite easy to swap based on identities:

r2 = r cos(θ) ⇒ (x2 + y2) = (x) ⇒ x2 + y2 = x

At this point, we're done: we've achieved a rectangular equation. However, if we want to understand it even better, we can go one more step and put this into a form we're used to-a circle:

x2 + y2 = x ⇒ x2 − x + y2 = 0 ⇒ x2 − x +

1

4

+ y2 =

1

4

⇒

⎛⎝

x−

1

2

⎞⎠

2

+ y2 =

⎛⎝

1

2

⎞⎠

2

If we're familiar with conic sections, we see that this is the graph of a circle centered at ([1/2], 0) with a radius of [1/2]: exactly what we would have gotten by graphing the original polar equation.

x2 + y2 = x, or, equivalently, (x−[1/2])2 + y2 = ([1/2] )2

Convert the equation from polar to rectangular, then solve for y.

r=

−1

1+sinθ

We can convert from polar coordinates to rectangular coordinates through the use of the following identities:

x = rcosθ, y = rsinθ, r2 = x2 + y2, tanθ =

y

x

When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.

Looking at the equation, we see that there's nothing we can currently swap out with an identity. However, having that fraction on the right side isn't helping things. Let's get rid of that by multiplying both sides by the denominator of 1+sinθ:

r=

−1

1+sinθ

⇒ r (1+sinθ) = −1 ⇒ r + r sinθ = −1

Now we see that we can apply the identity y = rsinθ:

r + r sinθ = −1 ⇒ r + y = −1

At this point, we still need to convert that r into something rectangular. The problem is that the only identity that uses just r is based on r2, which we don't currently have. However, we can get r2 to appear if we move the y to the other side, then square both sides:

r+ y = −1 ⇒ r = −1 − y ⇒ r2 = (−1−y)2

Now that r2 has appeared, we can apply the identity r2 = x2 + y2:

r2 = (−1−y)2 ⇒ x2 + y2 = (−1−y)2

We now have a rectangular equation, but the problem told us to also solve for y, so we need to get y alone on one side. Start off by expanding:

x2 + y2 = (−1−y)2 ⇒ x2 + y2 = 1 +2y + y2

We can subtract y2 on both sides, then get the y alone:

x2 + y2 = 1 +2y + y2 ⇒ x2 = 1+2y ⇒ x2 − 1 = 2y ⇒

1

2

x2 −

1

2

= y

y = [1/2] x2 − [1/2]

Convert the rectangular equation to polar form.

y = x

We can convert from polar coordinates to rectangular coordinates through the use of the following identities:

x = rcosθ, y = rsinθ, r2 = x2 + y2, tanθ =

y

x

When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.

We can apply the identities right from the start because they have x and y alone in them. Swap out based on the identities:

y = x ⇒ r sinθ = r cosθ

At this point, we've technically completed the problem, since we have an equation in polar form. Still, it might be nice to simplify it in case we wanted to graph it. In that case, start off by canceling out the r's on each side;

r sinθ = r cosθ ⇒ sinθ = cosθ

Next, we can combine the trig functions into a single tangent function by dividing cosθ on both sides:

sinθ = cosθ ⇒

sinθ

cosθ

= 1 ⇒ tanθ = 1

Finally, to make it super easy to graph, we can take the inverse tangent of both sides to just get θ alone:

tanθ = 1 ⇒ θ = tan−1 ( 1) ⇒ θ =

π

4

This makes sense! The rectangular equation y=x is just a straight line that goes up at an angle of 45° (because it has a slope of 1). The polar equation gives us the exact same line, just like it should.

θ = [(π)/4]

Convert the rectangular equation to polar form, then solve for r.

x2−9 = 6y

We can convert from polar coordinates to rectangular coordinates through the use of the following identities:

x = rcosθ, y = rsinθ, r2 = x2 + y2, tanθ =

y

x

When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.

This is a very tricky problem. Our knee-jerk reaction would be to swap out x and y based on the identities, as below:

x2−9 = 6y ⇒ (r cosθ)2 − 9 = 6 (rsinθ) ⇒ r2 cos2 θ− 9 = 6 ·r sinθ

However, it turns out that it's actually pretty difficult to solve for r from the above equation. Give it a try, and you'll quickly notice how tough it is to pin down r. [It is possible, but you have to use the identity sin2θ+ cos2 θ = 1 to change the cos2 to a sin2, then also do something similar to what we do below.] Instead, it will help us a bit to take a different approach we might not have initially thought of...

Notice that we have x2 on the left. One of the identities we have is r2 = x2 + y2, so alternatively, if we had y2 show up, we could use that identity. Let's try that instead:

x2 − 9 = 6y ⇒ x2 + y2 − 9 = y2 + 6y ⇒ r2 − 9 = y2 + 6y

At this point, we might try moving the 9 to the other side:

r2 − 9 = y2 + 6y ⇒ r2 = y2 + 6y + 9

Which then might cause us to realize we can factor it:

r2 = y2 + 6y + 9 ⇒ r2 = (y+3)2

Our goal is to eventually get r alone, so since we now have something squared on both sides, we might as well take the square root. Remember that taking a square root causes a `±' to show up.

r2 = (y+3)2 ⇒

√

r2

= ±

√

(y+3)2

⇒ r = ±(y+3)

We still need to fully convert to polar, so switch out the y:

r = ±(y+3) ⇒ r = ±(rsinθ+ 3)

Remember, ± means that there is both a + version and − version, so we can split the above into its two versions:

r = rsinθ+ 3

⎢⎢

r = −(rsinθ+3) ⇒ r = −rsinθ− 3

Finally, we have an equation that is entirely polar and it's not too difficult to solve for r. To do so, we need to get all the r's on one side, then pull out the r, then put everything else on the other side.

r=rsinθ+ 3

⎢⎢

r = −rsinθ− 3

r − r sinθ = 3

⎢⎢

r + r sinθ = −3

r(1−sinθ) = 3

⎢⎢

r(1+sinθ) = −3

r =

3

1−sinθ

⎢⎢

r =

−3

1+sinθ

At this point, we have a polar equation and we've solved for r. There is a little bit of strangeness, though... How can we have two different solutions? Each of those equations is distinct from the other, but somehow they are both conversions of the same rectangular equation? That's weird.

It turns out, there's an explanation. If you graph either of them, you get the exact same graph, which is pictured below. Each of the equations produces the same parabola that the original rectangular equation gave, they just do it from different starting locations. The graph of r = [3/(1−sinθ)] "starts" from the positive x-axis at 3, then works up to the right, until eventually flipping to the left side. The other one starts differently: r = [(−3)/(1+sinθ)] "starts" from the negative x-axis at −3, then works down at first, eventually coming up on the right side, and then also flipping like the other. The fact that we can express the original rectangular equation as two different polar equations is based on how we draw polar coordinates. We can name the same location with different sets of polar coordinates, and these two equations correspond to the different ways we can name the points. We can name with positive ("forward") r or we can name them with negative ("backward") r, which is the two types that come out of the equations.

r = [3/(1−sinθ)] or r = [(−3)/(1+sinθ)] [See the last step for a discussion of why the rectangular equation can be turned into two different polar equations.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Polar Equations & Functions

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